IGEM:IMPERIAL/2007/Experimental Design/Phase2/Protocol 1.2: Difference between revisions

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===Experiment 1===
==Experiment 1==
'''Status:'''
'''Status:'''
*Partially completed
*Partially completed
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*Gilson Pipette p5,p20,p200
*Gilson Pipette p5,p20,p200
'''Reagents'''
'''Reagents'''
*Recombinant GFP (Mw=27,000Da) [http://clontech.com/images/pacs/632373-PA16806.pdf Link to product information]. It is in the following form;
*Pure sample of recombinant GFP
*Pure sample of recombinant GFP
**100ug in buffer of 1mg/ml, this means in the solution there is:
**'''3.7x10<sup>-9</sup> moles''' of GFP
**'''0.037 M (moles dm<sup>-3</sup>)'''
**'''Volume supplied to us is 0.1mL'''
*DiiH<sub>2</sub>0
*DiiH<sub>2</sub>0
'''Protocol'''
'''Protocol'''
#Be careful about exposing the GFP samples to light to prevent photo bleaching and in addition kept on ice to prevent degradation
#Be careful about exposing the GFP samples to light to prevent photo bleaching and in addition kept on ice to prevent degradation
#Remove the sample of purified GFP from the feezer and place in a container at room temp, to allow to melt
#Remove the sample of purified GFP from the feezer and place in a container at room temp, to allow to melt
#Remove 10ul of purified GFP and add to 27ul of dii<sub>2</sub>0 in an PCR tube. This gives a 37ul of 10uM GFP concentration.
#Remove 15ul of purified GFP and add to a PCR tube. This give 6.7ul of 37uM GFP concentration.
#Remove 2ul of 10uM GFP to 18ul of dii<sub>2</sub>0 in an PCR tube.This gives a 10ul of 1uM GFP concentration
#Remove 7ul of purified GFP and add to 7ul of diiH<sub>2</sub>0 in an PCR tube. This gives a 14ul of 18uM GFP concentration
#Remove 8ul of 1uM GFP to 8ul of dii<sub>2</sub>0 in an PCR tube.This gives a 16ul of 0.5uM GFP concentration
#Remove 10ul of purified GFP and add to 27ul of diiH<sub>2</sub>0 in an PCR tube. This gives a 37ul of 10uM GFP concentration.
#Remove 8ul of 0.5uM GFP to 8ul of dii<sub>2</sub>0 in an PCR tube.This gives a 16ul of 0.0.25uM GFP concentration
#Remove 7ul of 10uM  GFP to 7ul of diiH<sub>2</sub>0 in an PCR tube. This gives a 37ul of 5uM GFP concentration.
#Remove 8ul of 0.25uM GFP to 8ul of dii<sub>2</sub>0 in an PCR tube.This gives a 16ul of 0.125uM GFP concentration
#Remove 2ul of 10uM GFP to 18ul of diiH<sub>2</sub>0 in an PCR tube.This gives a 10ul of 1uM GFP concentration
#Remove 8ul of 1uM GFP to 8ul of diiH<sub>2</sub>0 in an PCR tube.This gives a 16ul of 0.5uM GFP concentration
#Remove 8ul of 0.5uM GFP to 8ul of diiH<sub>2</sub>0 in an PCR tube.This gives a 16ul of 0.0.25uM GFP concentration
#Remove 8ul of 0.25uM GFP to 8ul of diiH<sub>2</sub>0 in an PCR tube.This gives a 16ul of 0.125uM GFP concentration
#Keep all samples on ice and in darkness
#Keep all samples on ice and in darkness


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'''Protocols'''
'''Protocols'''
#First collect all equipment and reagents and ensure that the fluorometer and that the PC connected has a data collection protocol installed.
#First collect all equipment and reagents and ensure that the fluorometer and that the PC connected has a data collection protocol installed.
#We will prepare 6xin vitro reactions for the GFP calibration curve
#We carried out x2 repeats for each concentration and so for each concentration we require the following mixtures:
#Commercial E.coli Cell Extract: First prepare a complete amino acid mixture for both extract solutions: Add the 15μl volume of two amino acid minus mixtures into an labeled eppendorf to give a volume of 30μl. Each amino acid minus mixture is missing one type of amino acid, and so by combining two solutions we are complementing each solution for the missing amino acid. Place eppendorf in a rack on bench.  
#Commercial E.coli Cell Extract: First prepare a complete amino acid mixture for both extract solutions: Add the 5μl volume of two amino acid minus mixtures into an labeled eppendorf to give a volume of 10μl. Each amino acid minus mixture is missing one type of amino acid, and so by combining two solutions we are complementing each solution for the missing amino acid. Place eppendorf in a rack on bench.  
#Commercial E.coli Cell Extract:Take an eppendorf tube and add 30µl of the E.coli complete amino acid mixture. Then add 120µl of S30 Premix Without Amino Acid. Then add 90µl of S30 Extract Circular. Add 15ul of dii water (This is to increase the volume to make the preparation of GFP dilutions easier. Place the eppendorf tube in a rack on the bench.  
#Commercial E.coli Cell Extract:Take an eppendorf tube and add 10µl of the E.coli complete amino acid mixture. Then add 40µl of S30 Premix Without Amino Acid. Then add 30 µl of S30 Extract Circular. Add 0.6ul of dii water (This is to increase the volume to make the preparation of GFP dilutions easier. Place the eppendorf tube in a rack on the bench.  
Vortex the tubes to mix thoroughly.  
Vortex the tubes to mix thoroughly.  
#Then to this eppendorf add 120ul of DNA of an empty plasmid
#Then to this eppendorf add 40ul of DNA of an empty plasmid
#Seperate out 60.3ul of the cell extract mixture into 6xeppendorf tubes.
#Seperate out 60.3ul of the cell extract mixture into 6xeppendorf tubes.
#To one tube add one of the 6.7ul of a GFP dilutions and for a control add 6.7 of diiwater. This means we are add 6.7ul of GFP to 67ul final volume, this gives a dilution factor of x10
#To one tube add one of the 6.7ul of a GFP dilutions and for a control add 6.7 of diiwater. This means we are add 6.7ul of GFP to 67ul final volume, this gives a dilution factor of x10 and so our final GFP concentrations are:
and so our final GFP concentrations are:
#*3.7uM
#*1.85uM
#*1uM
#*1uM
#*0.5uM
#*0.1uM
#*0.1uM
#*0.05uM
#*0.05uM
#*0.025uM
#*0.025uM
#*0.0125uM
#*0.0125uM
*Prepare the cell extract
*The moles of GFP that our in vitro systems is expected to produce is 7.4x10<sup>-12</sup> moles, this is based upon the quoted expected protein production of 200ng (2x10<sup>-7</sup>/27000Da). If the total volume of the cell extract is 100ul, this means that the concentration is around 7.4x10<sup>-8</sup> M.
*We need a sufficient range of dilutions to cover this range. However, the relationship between fluorescence and molcules of GFP is linear at concentrations in uM and lower. Therefore, if we have a sufficient range we should be able to plot an accurate calibration curve.
*We propose covering the following range of 8 data sets with 2 repeats for each:<br>
1.000uM<br>
0.500uM<br>
0.250uM<br>
0.125uM<br>
0.060uM<br>
'''Reagents'''
*Recombinant GFP (Mw=27,000Da) [http://clontech.com/images/pacs/632373-PA16806.pdf Link to product information]. It is in the following form;
*100ug in buffer of 1mg/ml, this means in the solution there is:
**'''3.7x10<sup>-9</sup> moles''' of GFP
**'''0.037 M (moles dm<sup>-3</sup>)'''
**'''Volume supplied to us is 0.1mL'''

Latest revision as of 07:03, 25 September 2007

Experiment 1

Status:

  • Partially completed
  • More Data Points needed

Aims:

  • To create and test a series of dilutions of GFP using a purified sample of GFP

Equipment

  • 5x PCR tubes 0.5ml
  • Gilson Pipette p5,p20,p200

Reagents

  • Recombinant GFP (Mw=27,000Da) Link to product information. It is in the following form;
  • Pure sample of recombinant GFP
    • 100ug in buffer of 1mg/ml, this means in the solution there is:
    • 3.7x10-9 moles of GFP
    • 0.037 M (moles dm-3)
    • Volume supplied to us is 0.1mL
  • DiiH20

Protocol

  1. Be careful about exposing the GFP samples to light to prevent photo bleaching and in addition kept on ice to prevent degradation
  2. Remove the sample of purified GFP from the feezer and place in a container at room temp, to allow to melt
  3. Remove 15ul of purified GFP and add to a PCR tube. This give 6.7ul of 37uM GFP concentration.
  4. Remove 7ul of purified GFP and add to 7ul of diiH20 in an PCR tube. This gives a 14ul of 18uM GFP concentration
  5. Remove 10ul of purified GFP and add to 27ul of diiH20 in an PCR tube. This gives a 37ul of 10uM GFP concentration.
  6. Remove 7ul of 10uM GFP to 7ul of diiH20 in an PCR tube. This gives a 37ul of 5uM GFP concentration.
  7. Remove 2ul of 10uM GFP to 18ul of diiH20 in an PCR tube.This gives a 10ul of 1uM GFP concentration
  8. Remove 8ul of 1uM GFP to 8ul of diiH20 in an PCR tube.This gives a 16ul of 0.5uM GFP concentration
  9. Remove 8ul of 0.5uM GFP to 8ul of diiH20 in an PCR tube.This gives a 16ul of 0.0.25uM GFP concentration
  10. Remove 8ul of 0.25uM GFP to 8ul of diiH20 in an PCR tube.This gives a 16ul of 0.125uM GFP concentration
  11. Keep all samples on ice and in darkness

Experiment 2

Status:

  • Partially completed
  • More Data Points needed

Aims:

  • To construct a calibration curve

Equipment

  • Fluorometer + Connected PC Turn on before beginning
  • 96 well plate x1 + Plate lid
  • 1.5ml eppendorf tube x7
  • eppendorf rack
  • Gilson p20,p200,p1000
  • Stop watch

Reagents

  • Our Prepared S30 extract
  • Commercial S30 E.coli extract. Including:
    • 175µl Amino Acid Mixture Minus Cysteine, 1mM
    • 175µl Amino Acid Mixture Minus Methionine, 1mM
    • 175µl Amino Acid Mixture Minus Leucine, 1mM
    • 450µl S30 Extract, Circular (3 × 150µl)
    • 750µl S30 Premix Without Amino Acids
  • Commercial S30 T7 extract. Including:
    • 175µl Amino Acid Mixture Minus Cysteine, 1mM
    • 175µl Amino Acid Mixture Minus Methionine, 1mM
    • 175µl Amino Acid Mixture Minus Leucine, 1mM
    • 450µl T7 S30 Extract, Circular (3 × 150µl)
    • 750µl S30 Premix Without Amino Acids
  • Dii water
  • GFP dilutions preapred above

Protocols

  1. First collect all equipment and reagents and ensure that the fluorometer and that the PC connected has a data collection protocol installed.
  2. We carried out x2 repeats for each concentration and so for each concentration we require the following mixtures:
  3. Commercial E.coli Cell Extract: First prepare a complete amino acid mixture for both extract solutions: Add the 5μl volume of two amino acid minus mixtures into an labeled eppendorf to give a volume of 10μl. Each amino acid minus mixture is missing one type of amino acid, and so by combining two solutions we are complementing each solution for the missing amino acid. Place eppendorf in a rack on bench.
  4. Commercial E.coli Cell Extract:Take an eppendorf tube and add 10µl of the E.coli complete amino acid mixture. Then add 40µl of S30 Premix Without Amino Acid. Then add 30 µl of S30 Extract Circular. Add 0.6ul of dii water (This is to increase the volume to make the preparation of GFP dilutions easier. Place the eppendorf tube in a rack on the bench.

Vortex the tubes to mix thoroughly.

  1. Then to this eppendorf add 40ul of DNA of an empty plasmid
  2. Seperate out 60.3ul of the cell extract mixture into 6xeppendorf tubes.
  3. To one tube add one of the 6.7ul of a GFP dilutions and for a control add 6.7 of diiwater. This means we are add 6.7ul of GFP to 67ul final volume, this gives a dilution factor of x10 and so our final GFP concentrations are:
    • 3.7uM
    • 1.85uM
    • 1uM
    • 0.5uM
    • 0.1uM
    • 0.05uM
    • 0.025uM
    • 0.0125uM
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